# ism5 - Chapter 5 Applications of Vectors in R2 and...

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Chapter 5 Applications of Vectors in R 2 and R 3 (Optional) Section 5.1, p. 263 2. (a) 4 i + j + 4 k . (b) 3 i 8 j k . (c) 0 i + 0 j + 0 k . (d) 4 i + 4 j + 8 k . 10. 1 2 90. 12. 1. T.1. (a) Interchange of the second and third rows of the determinant in (2) changes the sign of the determinant. (b) i j k u 1 u 2 u 3 v 1 + w 1 v 2 + w 2 v 3 + w 3 = i j k u 1 u 2 u 3 v 1 v 2 v 3 + i j k u 1 u 2 u 3 w 1 w 2 w 3 . (c) Similar to proof for (b). (d) Follows from the homogeneity property for determinants: Theorem 3.5. (e) Follows from Theorem 3.3. (f) Follows from Theorem 3.4. (g) First let u = i and verify that the result holds. Similarly, let u = j and then u = k . Finally, let u = u 1 i + u 2 j + u 3 k . (h) First let w = i . Then ( u × v ) × i = ( u 1 v 2 u 2 v 1 ) j ( u 3 v 1 u 1 v 3 ) k = u 1 ( v 1 i + v 2 j + v 3 k ) v 1 ( u 1 i + u 2 j + u 3 k ) = u 1 v v 1 u = ( i · u ) v ( i · v ) u . Thus equality holds when w = i . Similarly it holds when w = j , when w = k , and (adding scalar multiples of the three equations), when w = w 1 i + w 2 j + w 3 k . T.2. ( u × v ) · w = w 1 w 2 w 3 u 1 u 2 u 3 v 1 v 2 v 3 = u 1 u 2 u 3 v 1 v 2 v 3 w 1 w 2 w 3 = u · ( v × w )

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80 Chapter 5 T.3. We have j × i = i j k 0 1 0 1 0 0 = k 0 1 1 0 = k k × j = i j k 0 0 1 0 1 0 = i 0 1 1 0 = i i × k = i j k 1 0 0 0 0 1 = j 1 0 0 1 = j . T.4. ( u × v ) · w = ( u 2 v 3 u 3 v 2 ) w 1 + ( u 3 v 1 u 1 v 3 ) w 2 + ( u 1 v 2 u 2 v 1 ) w 3 = u 1 u 2 u 3 v 1 v 2 v 3 w 1 w 2 w 3 (expand the determinant along the third row).
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